Deriving the Area of a SphereDate: 10/21/2003 at 23:59:24 From: sana Subject: Deriving the area of sphere I know the area of a sphere is 4phi(r^2), but I'm wondering how to derive that formula. I know it should be done in cylindrical coordinates, and I'm thinking that the arc of a circle is defined as rd(theta) and it's multiplied with rd(phi) to get (r^2)d(theta)d(phi). Could you please help explain this? Date: 10/22/2003 at 02:19:09 From: Doctor Jeremiah Subject: Re: Deriving the area of sphere Hi Sana, I think the following answer from the Dr. Math archive will tell you what you want to know: Surface Area and Volume of a Sphere http://mathforum.org/library/drmath/view/55189.html Here is another answer not done in spherical coordinates--it's done by summing up thin circular sections: Deriving the Integral for the Surface Area of a Sphere http://mathforum.org/library/drmath/view/52133.html And here is one with spherical coordinates: Surface Area of a Sphere http://mathforum.org/library/drmath/view/51754.html And here is one that does it completely differently: Surface Area of a Sphere http://mathforum.org/library/drmath/view/51761.html I found these by searching our archives on these keywords: derive area sphere You can find other similar answers by doing the same search at: http://mathforum.org/library/drmath/mathgrepform.html I hope this helps. Let me know if you'd like to talk about this some more, or if you have any other questions. Be sure to search our archives when you have a question--you will often find your answer without having to write in and wait for a response. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 10/22/2003 at 03:25:42 From: Doctor Luis Subject: Re: Deriving the area of sphere Hi Sana, Please reference the diagram at the following url Spherical Coordinates http://mathworld.wolfram.com/SphericalCoordinates.html where theta is the azimuth angle (i.e. the angle measured from the x-axis if you project the point to the xy-plane), and phi is the polar angle (i.e. angle measured from the z-axis). The radius r is a constant for our sphere. The angle theta is also known as the longitude, and the angle phi is also known as the latitude. Imagine a little patch on the surface of the sphere, created by the differential angles d(theta) and d(phi). What is the area of that "square" patch? Now, be careful here. The side from the phi angle is easy, it's just r*d(phi), since it's just a small differential arc on a great circle on the sphere (if you look at the diagram, phi is measured on a great circle that has the z-axis as its diameter). The side from the theta angle is trickier. It's still a differential arc on a circle, so its length is still radius*d(theta), but the circle is smaller. You can see that on the diagram. The circle at the equator is largest (with radius r), but circles at higher latitudes have a smaller radius. You can see that these circles get smaller and smaller as you approach the poles. Do you see that in the diagram? Here is another diagram: As you can see, the radius of that circle is r*sin(phi). Therefore, the side of the patch from the d(theta) increment is (r*sin(phi))*d(theta). The area of the patch is then dA = [r d(phi)] * [r sin(phi) d(theta)] = r^2 sin(phi) d(phi) d(theta) To obtain the whole surface of the sphere, we integrate over all angles. The latitude phi ranges from 0 to pi, and the longitude theta ranges from 0 to 2pi. Therefore, our area is / / /2pi /pi | | | | A = | | |dA| = r^2 | d(theta) | sin(phi) d(phi) | | | | / / / 0 / 0 = r^2 * (2pi - 0) * (-cos(pi) + cos(0)) = r^2 * (2pi) * (2) = 4pi*r^2 and there you have it. Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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